POJ 2479 Maximum sum 解题报告
Maximum sum
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 40596 | Accepted: 12663 |
Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
Your task is to calculate d(A).
Input
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input
1 10 1 -1 2 2 3 -3 4 -4 5 -5
Sample Output
13
Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended.
Huge input,scanf is recommended.
Source
POJ Contest,Author:Mathematica@ZSU
题目大意:给定一个整数数字序列s,要求使两段不相交的子序列s1,s2的和最大
题解:动态规划
ls[i]表示以第i个元素结尾序列的最大值
rs[i]表示以第i个元素开始序列的最大值
rst[i]表示取i个元素能够达到的最大值
所以有
ls[i]=max(ls[i-1]+a[i], a[i]), rs[i]=max(rs[i+1]+a[i], a[i])
rst[i]=max(rst(i+1), rs[i])
所以最后的答案是s=max(s, ls[i]+rst(i+1))
#include <stdio.h> #include <string.h> #include <cstdio> #include <iostream> using namespace std; const int maxn = 1e6+7, inf = -1e6+7; int a[maxn], ls[maxn], rs[maxn], s, rst[maxn]; int main() { int t, n; scanf("%d", &t); while (t--) { s = 0; memset(ls, 0, sizeof(ls)); memset(rs, 0, sizeof(rs)); memset(rst, 0, sizeof(rst)); scanf("%d", &n); for (int i=0; i<n; i++) scanf("%d", &a[i]); ls[0] = a[0]; for (int i=1; i<n; i++) { ls[i] = max(ls[i-1]+a[i], a[i]); } rst[n-1] = rs[n-1] = a[n-1]; for (int i=n-2; i>=0; i--){ rs[i] = max(rs[i+1]+a[i], a[i]); rst[i] = max(rst[i+1], rs[i]); } s = inf; for (int i=0; i<n-1; i++) { s = max(s, ls[i]+rst[i+1]); } printf("%d\n", s); } return 0; }
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